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In: Physics

Two cars start from rest at a red stop light. When the light turns green, both...

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.6 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.1 seconds, before applying the brakes such that the car

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Expert Solution

For blue car ..
a) for initial acceleration
a = 3.6 m/sec
t = 4.4 sec
u = 0

so.. distance moved = s = ut + 0.5*a*t^2 = 0 + 0.5*3.6*4.4^2 = 34.848 m

speed after accelerationg = v = u + at = 0 + 3.6 * 4.4 = 15.84 m/sec

b) now for constant speed
speed = 15.84 m/sec
time = 8.1 sec

so. distance moved = 15.84 * 8.1 = 128.304 m

c) for retardation..

disance moved = s= 206.36 - 128.304 - 34.848 = 43.208 m
intial speed = u = 15.84 m/sec
final speed = v = 0
let the accletiaontion be a

so.. v^2 = u^2 + 2*a*s
0 = 15.84 ^2 + 2*a*43.208
so.. acceletiaon =a = - 2.90346 m/sec2

time of travel while retarding = ( v - u ) / a = ( 0 - 15.84) / (- 2.90346) = 5.45556 secs

4) so the acceletiaon of the blue car once the brake are applied = - 2.90346 m/sec2

5) so.. total time of travel for blue car = 5.45556 + 8.1 + 4.4 = 17.95556 secs..

6) for yellow.. let the acceeltaion be a ..
time t = 17.95556 secs
intial speed , u = 0
distance moved = s = 206.36 m

s = ut + 0.5*a*t^2
206.36 = 0 + 0.5 * a * 17.95556^2

so.. acceletiaon a = 1.28 m/sec2

genius_generous answered 1 hour ago
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